# Beginner abstract question. Why does a group operation of an element on itself return 1 or e?

I have some gaps in my understanding. This is my first abstract algebra course as an undergrad and the professor is truly all over the place. We are using Fraleigh's book and in the first week have covered pieces of the first 150 pages. Week 2 topics are Lagrange, cyclic, roots of unity, normal subgroups. A total mishmash as far as I’m concerned.

Ok, so my super basic question. Why does $$x\ast x$$ or $$x^2=1?$$ I understand $$xx^{-1}$$ would equal $$e=1$$.

The application. Prove that a group in which every element different from the identity element has order 2 is Abelian. The proof makes sense. The notion that $$x\ast x=1$$ does not.

• The idea that, for a group $G$, an element $x \in G$ has the property $x^2 = 1$ can be interpreted as every element $x$ being its own inverse. Does that answer your question? Sep 6, 2019 at 3:07
• Be sure to use your assumptions (what does order mean?) Sep 6, 2019 at 3:07
• "Prove that a group in which every element diff from the identity element that has order 2 is abelian." I find it difficult to parse this problem. Is this typed incorrectly? Sep 6, 2019 at 3:08
• If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group. My professor is Italian so he may have worded it strangely. Sep 6, 2019 at 4:50
• It might be worthwhile to point out that teaching, Lagrange theorem, cyclic groups, roots of unity and normal subgroups is not really a mish mash. This can flow very well from each other. Also these topis are very basic and as such seem like a reasonable second week for an abstract algebra class. You could really easilly fit them in one lecture.
– DRF
Sep 6, 2019 at 14:27

You've mixed yourself up. There are plenty of groups that have elements such that $$x^2 \neq 1$$. However, if you are dealing with a group such that $$x^2 = 1$$ for every element, then that group is Abelian.

• Fot the OP's understanding it might also help to note that in his exercise the fact that $x^2=1$ is an assumption of the proof. That really seems to be that place he's confused. (I realize you are trying to point that out with the bold if, but it might bear writing out more fully).
– DRF
Sep 6, 2019 at 14:34
• Yes, i totally missed that assumption. Thx Sep 6, 2019 at 19:34

Asserting that an element $$x$$ has order $$n$$ is the same thing as asserting that $$x^n=1$$ and that $$n$$ is the smallest natural number with this property.

So, in your problem every element $$x$$ other than $$1$$ has order $$2$$. And what this means is that precisely every element $$x$$ other than $$1$$ is such that $$x^2=1$$.

• .. and also $1$... Sep 6, 2019 at 3:31
• Well, yes, but see how I've put it. I wrote that “every element $x$ other than $1$ has order $2$” and, in the next sentence, I wrote what is it that this means. And so I had to repeat the “other than $1$” part. Sep 6, 2019 at 3:36
• Wow, great. Thx! Sep 6, 2019 at 4:51

Groups don't have to behave like integers. It's possible in a group to have $$x^2=e$$ for some particular few values of $$x$$ or even for all values of $$x$$. At the very least, $$e^2 =e$$. One of the requirements of a group, $$\mathbf G$$ is that, for every $$x$$ in $$\mathbf G$$, there must exists an element, designated $$x^{-1}$$, in $$\mathbf G$$ such that $$x x^{-1} = x^{-1} x = e$$. It is not required that $$x^2=e$$. It just happens some times.